# An object with a mass of 4 kg is acted on by two forces. The first is F_1= < 8 N , -6 N> and the second is F_2 = < 2 N, 7 N>. What is the object's rate and direction of acceleration?

Apr 7, 2017

The rate of acceleration is $2.5 \setminus {\text{m"/"s}}^{2}$ at the direction of ${5.7}^{\circ}$.

#### Explanation:

The question gives two forces in vector form.

The first step is to find the net force acting upon the object. This can be calculated by vector addition.

The sum of two vectors $< a , b >$ and $< c , d >$ is $< a + c , b + d >$.

Add the two force vectors $< 8 , - 6 >$ and $< 2 , 7 >$ to get $< 10 , 1 >$.

The next step is to find the magnitude of the vector, which is necessary to find the "size" of the force.

The magnitude of a vector $< a , b >$ is $\sqrt{{a}^{2} + {b}^{2}}$.

The "size" of the force is $\sqrt{{10}^{2} + {1}^{2}} = \sqrt{101} \setminus \text{N}$.

According to Newton's second law of motion, the net force acting upon an object is equal to the object's mass times its acceleration, or ${F}_{\text{net}} = m a$. The net force on the object is $\sqrt{101} \setminus \text{N}$, and its mass is $4 \setminus \text{kg}$. The acceleration is (sqrt(101)\ "N")/(4\ "kg")=sqrt(101)/4\ "m"/"s"^2~~2.5\ "m"/"s"^2.

Newton's first law of motion also states that the direction of acceleration is equal to the direction of its net force. The vector of its net force is $< 10 , 1 >$.

The angle "theta" of a vector $< a , b >$ is $\tan \left(\theta\right) = \frac{b}{a}$.

The angle $\theta$ of the direction of this vector is $\tan \left(\theta\right) = \frac{1}{10}$. Since both components of the vector are positive, the angle of the vector is in the first quadrant, or $0 < \theta < {90}^{\circ}$. Then, $\theta = \arctan \left(\frac{1}{10}\right) \approx {5.7}^{\circ}$ (the other possible value, ${185.7}^{\circ}$, is not correct since $0 < \theta < {90}^{\circ}$).